# Show the ring is a division algebra

Let $$R$$ be a commutative ring, and let $$M$$ be a simple $$R$$-module. Prove $$\text{End}_R(M)$$ is a division algebra.

Attempt: Since $$M$$ is simple for any $$\phi:M \rightarrow M$$, $$M/\{0\}\cong \phi(M)=M$$ by the first isomorphism theorem. Define a ring homomorphism $$\phi:R \rightarrow \text{End}_R(M)$$ such that $$\phi(R) \subset Z(\text{End}_R(M))$$.

Since $$M$$ is simple, all $$\alpha \in \text{End}_R(M)$$ must be bijective. So it is a division ring. Since $$M$$ is an $$R$$ module, the action of $$R$$ is defined by a ring homomorphism $$\sigma:R \rightarrow \text{End}_{Ab}(M)$$ So should I just define $$\phi(r)=\sigma(r)$$ for each $$r$$? Would $$\sigma(r)$$ commute with each element in $$\text{End}_R(M)$$ since $$R$$ is commutative? How do I know each $$\sigma(r) \in \text{End}_R(M)$$?

• I think that $\operatorname{End}_R(M)$ is an associative $R$-algebra is obvious. So only the bijection part has to be proven. But you are right, it is more a mention than a proof. Oct 13 at 21:40

There is an obvious map

$$\Phi: R \to \operatorname{End}_R(M)$$ defined by $$\Phi(r)(m) = rm.$$

Clearly $$\Phi(r)$$ is central for all $$r \in R$$, because if $$f \in \operatorname{End}_R(M)$$, then for all $$m\in M$$ $$(f\Phi(r))(m) = f(rm) = rf(m)= \Phi(r)(f(m)) = (\Phi(r)f)(m)$$ and thus $$\Phi(r) \in Z(\operatorname{End}_R(M))$$.

Using the ring homomorphism $$\Phi: R \to Z(\operatorname{End}_R(M))$$, the ring $$\operatorname{End}_R(M)$$ becomes an associative $$R$$-algebra.

You are seeing the concept of $$R$$-algebra incorrectly. I think the following equivalence will be helpful for you.

For a (not necessarily commutative) ring $$A$$ the following statements are equivalent:

• There is a ring homomorphism $$R \to A$$ whose image lies in $$Z(A)$$.
• There is an $$R$$-module structure on $$A$$ such that $$(rx)y = r(xy) \quad\&\quad x(ry) = r(xy) \tag{1}$$ for all $$r \in R$$ and $$x,y \in A$$ (equivalently, the product $$A \times A \to A,(x,y) \mapsto xy$$ is $$R$$-bilinear).

Indeed, if there is a ring homomorphism $$\nu : R \to A$$, note that $$A$$ (being an $$A$$-module) gives us a ring homomorphism $$\mu : A \to \operatorname{End}_\textsf{Ab}(A)$$, and then $$\mu \circ \nu$$ yields an $$R$$-module structure to $$A$$. Explicitly, for $$r \in R$$ and $$a \in A$$, $$ra := \nu(r)a$$ (the product in $$A$$ of $$\nu(r)$$ with $$a$$). The equality $$(1)$$ is true if the image of $$\nu$$ lies in $$Z(A)$$.

Conversely, if there is an $$R$$-module structure on $$A$$ such that $$(1)$$ holds, then for each $$r \in R$$ and $$a \in A$$, $$(r1_A)a = r(1_Aa) = ra \quad\&\quad a(r1_A) = r(a1_A) = ra$$ so $$(r1_A)a = a(r1_A)$$. Hence, the image of the map $$\_1_A : R \to A$$, $$r \mapsto r1_A$$ is inside of $$Z(A)$$. Finally, from $$(1)$$ and the commutativity of $$R$$ one can also deduce that $$(r_11_A)(r_21_A) = (r_1r_2)1_A$$ for all $$r_1,r_2 \in R$$, meaning that $$\_1_A$$ is a ring homomorphism.

Thus, once is clear how $$\operatorname{End}_R(M)$$ is an $$R$$-algebra, it remains to be seen how the simplicity of $$M$$ implies that each nonzero $$f \in \operatorname{End}_R(M)$$ is invertible, but this is easy because $$f \neq 0$$ implies that $$\ker f \neq M$$ and $$\operatorname{im} f \neq \{0\}$$.